linux - how to use exec() without any parameter? -


i wanna run "ls" command exec(), , code exec("/bin/ls", null) got text shows "a null argv[0] passed through exec system call." if add "all" parameter woks. exec("/bin/ls","all",null)

however, when use exec("/bin/ps", null), works properly. me figure out whats wrong program?

btw: use execl()

#include <iostream> #include <unistd.h>             //required fork() #include <sys/types.h>          //required wait() #include <sys/wait.h>           //required wait()  using namespace std;  int main(){         string cmd="";         string cmdpath="/bin/";         cout<<endl<<getcwd(null,0)<<" >> ";         cin>>cmd;         cout<<endl;         string cmdcmdpath = cmdpath+cmd;         const char* charcmd = cmdcmdpath.c_str();          int x = fork();         if(x!=0){                 cout<<"the command "<<cmd<<" running"<<endl;                 wait(null);                 cout<<"im parent!"<<endl;         }else if (x==0){                 cout<<"im child!"<<endl;                 execl(charcmd,null);                 cout<<"child done"<<endl;         }  }

carefully read desctiption of execl:

the first argument, convention, should point filename associated file being executed.

it means, second execl parameter should path, referred same file first one. usually, first , second parameters same:

execl(charcmd, charcmd, null);

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